/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could be proven: (0) OTRS (1) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) MRRProof [EQUIVALENT, 13 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 23 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: f(x, x) -> g(f(x, x)) g(x) -> a Outermost Strategy. ---------------------------------------- (1) Thiemann-SpecialC-Transformation (EQUIVALENT) We applied the Thiemann-SpecialC transformation to transform the outermost TRS to an innermost TRS. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_f(x, x) -> result_f(g(f(x, x))) redex_g(x) -> result_g(a) check_f(result_f(x)) -> go_up(x) check_g(result_g(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) check_g(redex_g(x_1)) -> in_g_1(reduce(x_1)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_g_1(go_up(x_1)) -> go_up(g(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) TOP(go_up(x)) -> REDUCE(x) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) REDUCE(f(x_1, x_2)) -> REDEX_F(x_1, x_2) REDUCE(g(x_1)) -> CHECK_G(redex_g(x_1)) REDUCE(g(x_1)) -> REDEX_G(x_1) CHECK_F(redex_f(x_1, x_2)) -> IN_F_1(reduce(x_1), x_2) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2)) -> IN_F_2(x_1, reduce(x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) CHECK_G(redex_g(x_1)) -> IN_G_1(reduce(x_1)) CHECK_G(redex_g(x_1)) -> REDUCE(x_1) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_f(x, x) -> result_f(g(f(x, x))) redex_g(x) -> result_g(a) check_f(result_f(x)) -> go_up(x) check_g(result_g(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) check_g(redex_g(x_1)) -> in_g_1(reduce(x_1)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_g_1(go_up(x_1)) -> go_up(g(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_f(x, x) -> result_f(g(f(x, x))) redex_g(x) -> result_g(a) check_f(result_f(x)) -> go_up(x) check_g(result_g(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) check_g(redex_g(x_1)) -> in_g_1(reduce(x_1)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_g_1(go_up(x_1)) -> go_up(g(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: redex_f(x, x) -> result_f(g(f(x, x))) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: redex_f(x, x) -> result_f(g(f(x, x))) The set Q consists of the following terms: redex_f(x0, x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) Used ordering: Polynomial interpretation [POLO]: POL(CHECK_F(x_1)) = 1 + x_1 POL(REDUCE(x_1)) = 2*x_1 POL(f(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(g(x_1)) = x_1 POL(redex_f(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(result_f(x_1)) = x_1 ---------------------------------------- (13) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: redex_f(x, x) -> result_f(g(f(x, x))) The set Q consists of the following terms: redex_f(x0, x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_f(x, x) -> result_f(g(f(x, x))) redex_g(x) -> result_g(a) check_f(result_f(x)) -> go_up(x) check_g(result_g(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) check_g(redex_g(x_1)) -> in_g_1(reduce(x_1)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_g_1(go_up(x_1)) -> go_up(g(x_1)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_g(x) -> result_g(a) check_g(result_g(x)) -> go_up(x) redex_f(x, x) -> result_f(g(f(x, x))) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) in_g_1(go_up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(go_up(x0)) in_g_1(go_up(x0)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_g(x) -> result_g(a) check_g(result_g(x)) -> go_up(x) redex_f(x, x) -> result_f(g(f(x, x))) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. TOP(go_up(x)) -> TOP(reduce(x)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO]: Non-tuple symbols: <<< M( result_f_1(x_1) ) = [[0], [0]] + [[1, 1], [0, 0]] * x_1 >>> <<< M( a ) = [[0], [0]] >>> <<< M( reduce_1(x_1) ) = [[1], [0]] + [[0, 0], [1, 1]] * x_1 >>> <<< M( in_f_2_2(x_1, x_2) ) = [[0], [1]] + [[0, 0], [1, 1]] * x_1 + [[1, 0], [1, 1]] * x_2 >>> <<< M( redex_f_2(x_1, x_2) ) = [[1], [0]] + [[1, 1], [0, 0]] * x_1 + [[1, 1], [0, 0]] * x_2 >>> <<< M( check_f_1(x_1) ) = [[1], [1]] + [[0, 0], [1, 0]] * x_1 >>> <<< M( g_1(x_1) ) = [[1], [0]] + [[0, 0], [0, 0]] * x_1 >>> <<< M( go_up_1(x_1) ) = [[1], [1]] + [[0, 0], [1, 1]] * x_1 >>> <<< M( result_g_1(x_1) ) = [[0], [1]] + [[0, 0], [1, 1]] * x_1 >>> <<< M( in_f_1_2(x_1, x_2) ) = [[0], [1]] + [[1, 0], [1, 1]] * x_1 + [[0, 0], [1, 1]] * x_2 >>> <<< M( f_2(x_1, x_2) ) = [[1], [1]] + [[0, 0], [1, 1]] * x_1 + [[1, 0], [0, 1]] * x_2 >>> <<< M( check_g_1(x_1) ) = [[1], [0]] + [[0, 0], [0, 1]] * x_1 >>> <<< M( redex_g_1(x_1) ) = [[0], [1]] + [[0, 0], [0, 0]] * x_1 >>> Tuple symbols: <<< M( TOP_1(x_1) ) = [[0]] + [[0, 1]] * x_1 >>> Matrix type: We used a basic matrix type which is not further parametrizeable. As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order. The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) redex_f(x, x) -> result_f(g(f(x, x))) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) redex_g(x) -> result_g(a) check_g(result_g(x)) -> go_up(x) ---------------------------------------- (22) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) reduce(g(x_1)) -> check_g(redex_g(x_1)) redex_g(x) -> result_g(a) check_g(result_g(x)) -> go_up(x) redex_f(x, x) -> result_f(g(f(x, x))) check_f(result_f(x)) -> go_up(x) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: reduce(f(x0, x1)) reduce(g(x0)) redex_f(x0, x0) redex_g(x0) check_f(result_f(x0)) check_g(result_g(x0)) check_f(redex_f(x0, x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (24) YES